1
$\begingroup$

I asked a similar question here more than two years ago. I did not get an answer to my complete satisfaction. I would like to reiterate the problem again.

The local electric field of a monochromatic radiation is nonzero and varies sinusoidally in a predictable fashion. For example, the electric field of an ideal monochromatic radiation is described by $${\bf E}({\bf r},t)={E}_0\hat{{ \varepsilon}}\cos({\bf k}\cdot{\bf r}-\omega t),$$ at any location ${\bf r}$, is nonzero at any time $t$ and varies with time in a predictable manner. Here, $E_0$ is a fixed number and so is $\omega$ (the frequency of the radiation), and $\hat{\varepsilon}$ represents the constant polarization vector.

In contrast, assuming that the electric field at any location of a Blackbody cavity is due to an incoherent superposition of electric fields of all frequencies, polarizations (and all amplitudes?), can we rigorously come up with a mathematical expression for the local electric field at any time $t$ for the blackbody radiation? I am interested in getting a mathematical formula that properly represents the local electric field of incoherent thermal radiation.

$\endgroup$
2
  • $\begingroup$ Would an expression for the expectation value of $\mathbf{E}(\mathbf{r})$ be sufficient? I doubt that you can get the exact expression for $\mathbf{E}(\mathbf{r})$ itself; we never really know the microstate of a system in thermal equilibrium (if we did, we wouldn't be using thermodynamics to describe it.) $\endgroup$ Aug 31 '20 at 13:52
  • $\begingroup$ Do you mean a thermal average? I think that will do. $\endgroup$
    – SRS
    Aug 31 '20 at 14:58
0
$\begingroup$

Using Cartesian coordinates you could write something like: \begin{eqnarray} \vec{E}(\vec{r},t) & = & \int E_{0}(\omega) \left(\cos (\omega[t - x/c]) \hat{j} + \cos(\omega[t -x/c] +\phi_{x,\omega}) \hat{k}\right)\ d\omega \nonumber \\ & + & \int E_0(\omega)\left( \cos (\omega[t - y/c]) \hat{i} + \cos(\omega[t -y/c] +\phi_{y,\omega}) \hat{k}\right)\ d\omega \nonumber \\ & + & \int E_0(\omega)\left( \cos (\omega[t - z/c]) \hat{i} + \cos(\omega[t -z/c] +\phi_{z, \omega}) \hat{j}\right)\ d\omega . \nonumber \end{eqnarray} This represents the sum of three unpolarised beams of light travelling in three directions, each with equal power, summed over all frequencies. In each term, the $\phi_{i,\omega}$ term represents a random phase between $0$ and $2\pi$.

The time averaged value of $\vec{E}\cdot \vec{E}$ is $$\left< \vec{E}\cdot \vec{E}\right> = 3 \int E_0^2(\omega)\ d\omega $$ and so the time-averaged energy density of the electromagnetic fields would be $$ 6\epsilon_0 \int E_0^2(\omega)\ d\omega = \int \frac{\hbar \omega^3}{\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right)\ d\omega\ , $$ where the term on the right is the total energy density of a blackbody radiation field and the extra factor of 2 on the left hand side accounts for equal energy density in the magnetic field. Thus $$E_0(\omega) = \left[ \frac{\hbar \omega^3}{6\epsilon_0\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right) \right]^{1/2} $$

Of course, the time-average of the electric field itself is zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.